Question

What is `lim_(nrarroo)(-1)^(n-1)sin(pisqrt(n^2+0.5n+1))` ?

Answer 1

The limit is `-1/sqrt2`

Explanation:

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We need to find the asymptote to `sqrt(n^2+0.5n+1)`.
(Thanks, George C.)

To find (linear) oblique asymptote `y=mn+b`,

`m = lim_(nrarroo) sqrt(n^2+0.5n+1)/n = 1`.

Now we find `b`:

`b= lim_(nrarroo) sqrt(n^2+0.5n+1)/mn = lim_(nrarroo) sqrt(n^2+0.5n+1)/n = 1/4`.

So, `sqrt(n^2+0.5n+1)` is asymptotic to `n+1/4`.

Alternative method from George C

`sqrt(n^2+0.5n+1) = sqrt((n+1/4)^2+15/16)`

`= (n+1/4)sqrt(1+15/(16(n+1/4)^2)`

(If I've deduced the method correctly, our goal is to get `(mn+b)sqrt(1+g(n))` where `lim_(nrarroo)g(n)=0`.)

Returning to the big question

`lim_(nrarroo)(-1)^(n-1)sin(pisqrt(n^2+0.5n+1)) = lim_(nrarroo)(-1)^(n-1)sin(pi(n+1/4)sqrt(1+15/(16(n+1/4)^2)))`

For even `n`, say `n=2k`,
we get `-sin(2kpi+pi/4) = -sin(pi/4) = -1/sqrt2`

For odd `n`, say `n=2k+1`,
we get `sin(2kpi+pi+pi/4) = sin((5pi)/4) = -1/sqrt2`

Therefore,

`lim_(nrarroo)(-1)^(n-1)sin(pisqrt(n^2+0.5n+1)) = -1/sqrt2`

Answer 2

`-sqrt2/2`

Explanation:

Extra care should be taken since we are dealing with a periodic function of the kind
`f(x)=+-sinx`

It's not sufficient that, when `n->oo`, `n^2` is much greater than `0.5n+1`, because if `sqrt(n^2+0.5n+1)-n > 0` than the function will have a value possibly different than zero.

Intuitively it's easy to see that `sqrt(k+1)->sqrt k` (or, even more importantly for the case, that `sqrt(k+1)-sqrt k->0`) when `k->oo`
Or
`sqrt(1001)-sqrt(1000)=0.015,807`
and
`sqrt(1,000,001)-sqrt(1,000,000)=0.000,500`
and so on.

But let's prove that `sqrt(n^2+0.5n+a)->sqrt(n^2+0.5n)` when `n->oo`, where `n in NN` and `a in RR`
`lim_(n->oo) (sqrt(n^2+0.5n+a)-sqrt(n^2+0.5n))*((sqrt(n^2+0.5n+a)+sqrt(n^2+0.5n)))/(sqrt(n^2+0.5n+a)+sqrt(n^2+0.5n))=`
`=lim_(n->oo) (cancel(n^2)+cancel(0.5n)+a-cancel(n^2)-cancel(0.5n))/(sqrt(n^2+0.5n+a)+sqrt(n^2+0.5n))`
`=lim_(n->oo) a/(sqrt(n^2+0.5n+a)+sqrt(n^2+0.5n))`
`=lim_(n->oo) a/oo=0`

Since the value of `a` (in the present case `a=1`) doesn't matter to the result, we can make
`lim_(n->oo) sqrt(n^2+0.5n+1)=lim_(n->oo) sqrt(n^2+0.5n+1/16)=lim_(n->oo) sqrt((n+1/4)^2)=lim_(n->oo) (n+1/4)`

So the main expression becomes

`=lim_(n->oo) (-1)^(n-1)sin[pi(n+1/4)]`
`=lim_(n->oo) (-1)^(n-1)sin(n*pi+pi/4)`
But
`sin(n*pi+pi/4)=sin(n*pi)*cos(pi/4)+sin(pi/4)*cos(n.pi)=`
(remembering that `n in NN` => `sin(n*pi)=0`)
`=sqrt2/2*cos(n*pi)`
In the main expression
`=lim_(n->oo) (-1)^(n-1)*sqrt2/2*cos(n*pi)`

Now consider the possible values of the expression above
If n is odd `=> (-1)^("even number").sqrt2/2*cos[("odd number")*pi]=1*sqrt2/2(-1)=-sqrt2/2`
If n is even `=> (-1)^("odd number").sqrt2/2*cos[("even number")*pi]=(-1)*sqrt2/2*1=-sqrt2/2`

So, for `n in NN`
`=lim_(n->oo) (-1)^(n-1)*sqrt2/2*cos(n*pi)=-sqrt2/2`

Answer 3

`-1/sqrt2`.

Explanation:

I have revised my answer, thanks to George for his affirmation on his answer. I am sorry for having overlooked some nicety in applying the notion of 'one is asymptotic with another', in this limit problem.

The correction follows, without changing my approach.

`pisqrt(n^2+0.5n+1)=npisqrt(1+0.5/n+1/n^2))=npi(1+(1/2)(0.5/n)+O(1/n^2))`, using expansion in powers of `1/n.`
Here, `ntooo` through integer values 1,2,3,..,

The limit is same as the limit for
`(-1)^nsin(npi+pi/4+O(1/n)) =-1/sqrt2`, as given in the other answers..